CAT 2020 Quant Question [Slot 3] with Solution 11

Since the roots are real \({m^2} - 8n \ge 0\) and \({(2n)^2} - 4m \ge 0\) \( \Rightarrow {n^2} - m \ge 0\)

\( \Rightarrow {n^4} \ge {m^2} \ge 8n\)

\( \Rightarrow n \ge 2\) and \(m \ge 4\)

Hence the least value of \(m + n = 2 + 4 = 6\)